What is the integral from a to b of 1/(b-a)x ?

The integral from a to b of 1/(b-a)x is (b+a)/2

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integral from a to b of 1/(b-a)x

\int_{a}^{b} \frac{1}{b - a} x d x

\frac{b + a}{2}

Solution steps

\int_{a}^{b} \frac{1}{b - a} x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{b - a} \cdot \int_{a}^{b} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{a}^{b}

= \frac{1}{b - a} [\frac{x ^{2}}{2}]_{a}^{b}

Simplify

\frac{1}{b - a} [\frac{x ^{2}}{2}]_{a}^{b} : \frac{[ \frac{x ^{2}}{2} ] _{a}^{b}}{b - a}

= \frac{[ \frac{x ^{2}}{2} ] _{a}^{b}}{b - a}

Compute the boundaries:

\frac{b ^{2}}{2} - \frac{a ^{2}}{2}

= \frac{\frac{b ^{2}}{2} - \frac{a ^{2}}{2}}{b - a}

Simplify

= \frac{b + a}{2}

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y = e^{x}, y = 7 e^{- x}, x = 0

\int ln (x + x^{2} - 1) d x

What is the integral from a to b of 1/(b-a)x ?
The integral from a to b of 1/(b-a)x is (b+a)/2